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Question
A sequence b0, b1, b2 ... is defined by letting b0 = 5 and bk = 4 + bk – 1 for all natural numbers k. Show that bn = 5 + 4n for all natural number n using mathematical induction.
Solution
We have b0 = 5 and bk = 4 + bk – 1
⇒ b0 = 5, b1 = 4 + b0 = 4 + 5 = 9
And b2 = 4 + b1 = 4 + 9 = 13
Let P(n) : bn = 5 + 4n
Step 1: P(1) : b1 = 5 + 4 = 9
⇒ 9 = 9 which is true.
Step 2: P(k): bk = 5 + 4k
Let it be true ∀ k ∈ N.
Step 3: Given that P(k) = 4 + bk – 1
⇒ P(k + 1) = 4 + bk+1–1
⇒ P(k + 1) = 4 + bk = 4 + 5 + 4k
⇒ P(k + 1) = 5 + 4(k + 1) which is true for P(k + 1).
Hence, P(k + 1) is true whenever P(k) is true.
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